165700
Two capacitors, one $4 \mathrm{pF}$ and the other $6 \mathrm{pF}$, connected in parallel, are charged by a $100 \mathrm{~V}$ battery. The energy stored in the capacitors is
1 $12 \times 10^{-8} \mathrm{~J}$
2 $2.4 \times 10^{-8} \mathrm{~J}$
3 $5.0 \times 10^{-8} \mathrm{~J}$
4 $1.2 \times 10^{-6} \mathrm{~J}$
Explanation:
: Given, $\mathrm{C}_{1}=4 \mathrm{pF}$ and $\mathrm{C}_{2}=6 \mathrm{pF}$ $\text { Voltage }=100 \mathrm{~V}$ Both capacitors connected in parallel. Hence, equivalent capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=4 \mathrm{pF}+6 \mathrm{pF}=10 \mathrm{pF}$ $\mathrm{C}=10 \times 10^{-12} \mathrm{~F}$ Energy stored in the capacitor $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times(100)^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times 10^{4}$ $\mathrm{E} =5.0 \times 10^{-8} \mathrm{~J}$
[Manipal UGET-2017]
Capacitance
165701
Two conductors of the same material have their diameters in the ratio $1: 2$ and their lengths in the ratio $2: 1$. If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be
165702
As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$, then energy given by the battery will be
1 $\frac{1}{2} \mathrm{CV}^{2}$
2 more than $\frac{1}{2} \mathrm{CV}^{2}$
3 less than $\frac{1}{2} \mathrm{CV}^{2}$
4 zero
Explanation:
: We know, Energy stored in a capacitor $\mathrm{E}=\frac{1}{2} \mathrm{CV}_{\mathrm{C}}^{2}$ Where, $\mathrm{V}_{\mathrm{C}}=$ Voltage drop across capacitor. Apply KVL in Circuit $\mathrm{V}=\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{C}}$ We know, in steady state current through resistance $=0$ $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}=0$ Hence, $\mathrm{V}=\mathrm{V}_{\mathrm{C}}$ From equation (i) $\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}$
[CG PET -2018]
Capacitance
165703
The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by
: We know, Total energy Stored in a capacitor, $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $\text { or } \quad \mathrm{E} =\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}^{2} \quad\left(\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right)$ Volume between the plate $=\mathrm{A} \times \mathrm{d}$ Hence, Energy per unit Volume, $\frac{E}{A d}=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right) V^{2}}{A d}=\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}$
[CG PET- 2017
Capacitance
165704
A $600 \mathrm{pF}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is
1 $6 \times 10^{-6} \mathrm{~J}$
2 $3 \times 10^{-6} \mathrm{~J}$
3 $6 \times 10^{-9} \mathrm{~J}$
4 $3 \times 10^{-9} \mathrm{~J}$
Explanation:
: Given that, Capacitance of Capacitor $(\mathrm{C})=600 \mathrm{pF}=600 \times 10^{-12} \mathrm{~F}$ and potential difference $(\mathrm{V})=200 \mathrm{~V}$ Electrostatic Energy stored $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=\frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^{4}$ $=1.2 \times 10^{-5} \mathrm{~J}$ After disconnecting, A $600 \mathrm{pF}$ capacitor added. Then equivalent capacitance $\left(\mathrm{C}^{\prime}\right)$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $=\frac{1}{600}+\frac{1}{600}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{300}$ $\mathrm{C}^{\prime} =300 \mathrm{pF}$ Then, new electrostatic energy, $\mathrm{E}^{\prime} =\frac{1}{2} \mathrm{C}^{\prime} \times \mathrm{V}^{2}$ $=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2}$ $=0.6 \times 10^{-5} \mathrm{~J}$ Loss in electrostatic energy $\left(\mathrm{E}-\mathrm{E}^{\prime}\right)$ $=1.2 \times 10^{-5}-0.6 \times 10^{5}$ $=0.6 \times 10^{-5}$ $=6 \times 10^{-6} \mathrm{~J}$
165700
Two capacitors, one $4 \mathrm{pF}$ and the other $6 \mathrm{pF}$, connected in parallel, are charged by a $100 \mathrm{~V}$ battery. The energy stored in the capacitors is
1 $12 \times 10^{-8} \mathrm{~J}$
2 $2.4 \times 10^{-8} \mathrm{~J}$
3 $5.0 \times 10^{-8} \mathrm{~J}$
4 $1.2 \times 10^{-6} \mathrm{~J}$
Explanation:
: Given, $\mathrm{C}_{1}=4 \mathrm{pF}$ and $\mathrm{C}_{2}=6 \mathrm{pF}$ $\text { Voltage }=100 \mathrm{~V}$ Both capacitors connected in parallel. Hence, equivalent capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=4 \mathrm{pF}+6 \mathrm{pF}=10 \mathrm{pF}$ $\mathrm{C}=10 \times 10^{-12} \mathrm{~F}$ Energy stored in the capacitor $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times(100)^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times 10^{4}$ $\mathrm{E} =5.0 \times 10^{-8} \mathrm{~J}$
[Manipal UGET-2017]
Capacitance
165701
Two conductors of the same material have their diameters in the ratio $1: 2$ and their lengths in the ratio $2: 1$. If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be
165702
As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$, then energy given by the battery will be
1 $\frac{1}{2} \mathrm{CV}^{2}$
2 more than $\frac{1}{2} \mathrm{CV}^{2}$
3 less than $\frac{1}{2} \mathrm{CV}^{2}$
4 zero
Explanation:
: We know, Energy stored in a capacitor $\mathrm{E}=\frac{1}{2} \mathrm{CV}_{\mathrm{C}}^{2}$ Where, $\mathrm{V}_{\mathrm{C}}=$ Voltage drop across capacitor. Apply KVL in Circuit $\mathrm{V}=\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{C}}$ We know, in steady state current through resistance $=0$ $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}=0$ Hence, $\mathrm{V}=\mathrm{V}_{\mathrm{C}}$ From equation (i) $\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}$
[CG PET -2018]
Capacitance
165703
The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by
: We know, Total energy Stored in a capacitor, $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $\text { or } \quad \mathrm{E} =\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}^{2} \quad\left(\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right)$ Volume between the plate $=\mathrm{A} \times \mathrm{d}$ Hence, Energy per unit Volume, $\frac{E}{A d}=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right) V^{2}}{A d}=\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}$
[CG PET- 2017
Capacitance
165704
A $600 \mathrm{pF}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is
1 $6 \times 10^{-6} \mathrm{~J}$
2 $3 \times 10^{-6} \mathrm{~J}$
3 $6 \times 10^{-9} \mathrm{~J}$
4 $3 \times 10^{-9} \mathrm{~J}$
Explanation:
: Given that, Capacitance of Capacitor $(\mathrm{C})=600 \mathrm{pF}=600 \times 10^{-12} \mathrm{~F}$ and potential difference $(\mathrm{V})=200 \mathrm{~V}$ Electrostatic Energy stored $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=\frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^{4}$ $=1.2 \times 10^{-5} \mathrm{~J}$ After disconnecting, A $600 \mathrm{pF}$ capacitor added. Then equivalent capacitance $\left(\mathrm{C}^{\prime}\right)$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $=\frac{1}{600}+\frac{1}{600}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{300}$ $\mathrm{C}^{\prime} =300 \mathrm{pF}$ Then, new electrostatic energy, $\mathrm{E}^{\prime} =\frac{1}{2} \mathrm{C}^{\prime} \times \mathrm{V}^{2}$ $=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2}$ $=0.6 \times 10^{-5} \mathrm{~J}$ Loss in electrostatic energy $\left(\mathrm{E}-\mathrm{E}^{\prime}\right)$ $=1.2 \times 10^{-5}-0.6 \times 10^{5}$ $=0.6 \times 10^{-5}$ $=6 \times 10^{-6} \mathrm{~J}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Capacitance
165700
Two capacitors, one $4 \mathrm{pF}$ and the other $6 \mathrm{pF}$, connected in parallel, are charged by a $100 \mathrm{~V}$ battery. The energy stored in the capacitors is
1 $12 \times 10^{-8} \mathrm{~J}$
2 $2.4 \times 10^{-8} \mathrm{~J}$
3 $5.0 \times 10^{-8} \mathrm{~J}$
4 $1.2 \times 10^{-6} \mathrm{~J}$
Explanation:
: Given, $\mathrm{C}_{1}=4 \mathrm{pF}$ and $\mathrm{C}_{2}=6 \mathrm{pF}$ $\text { Voltage }=100 \mathrm{~V}$ Both capacitors connected in parallel. Hence, equivalent capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=4 \mathrm{pF}+6 \mathrm{pF}=10 \mathrm{pF}$ $\mathrm{C}=10 \times 10^{-12} \mathrm{~F}$ Energy stored in the capacitor $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times(100)^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times 10^{4}$ $\mathrm{E} =5.0 \times 10^{-8} \mathrm{~J}$
[Manipal UGET-2017]
Capacitance
165701
Two conductors of the same material have their diameters in the ratio $1: 2$ and their lengths in the ratio $2: 1$. If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be
165702
As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$, then energy given by the battery will be
1 $\frac{1}{2} \mathrm{CV}^{2}$
2 more than $\frac{1}{2} \mathrm{CV}^{2}$
3 less than $\frac{1}{2} \mathrm{CV}^{2}$
4 zero
Explanation:
: We know, Energy stored in a capacitor $\mathrm{E}=\frac{1}{2} \mathrm{CV}_{\mathrm{C}}^{2}$ Where, $\mathrm{V}_{\mathrm{C}}=$ Voltage drop across capacitor. Apply KVL in Circuit $\mathrm{V}=\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{C}}$ We know, in steady state current through resistance $=0$ $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}=0$ Hence, $\mathrm{V}=\mathrm{V}_{\mathrm{C}}$ From equation (i) $\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}$
[CG PET -2018]
Capacitance
165703
The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by
: We know, Total energy Stored in a capacitor, $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $\text { or } \quad \mathrm{E} =\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}^{2} \quad\left(\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right)$ Volume between the plate $=\mathrm{A} \times \mathrm{d}$ Hence, Energy per unit Volume, $\frac{E}{A d}=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right) V^{2}}{A d}=\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}$
[CG PET- 2017
Capacitance
165704
A $600 \mathrm{pF}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is
1 $6 \times 10^{-6} \mathrm{~J}$
2 $3 \times 10^{-6} \mathrm{~J}$
3 $6 \times 10^{-9} \mathrm{~J}$
4 $3 \times 10^{-9} \mathrm{~J}$
Explanation:
: Given that, Capacitance of Capacitor $(\mathrm{C})=600 \mathrm{pF}=600 \times 10^{-12} \mathrm{~F}$ and potential difference $(\mathrm{V})=200 \mathrm{~V}$ Electrostatic Energy stored $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=\frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^{4}$ $=1.2 \times 10^{-5} \mathrm{~J}$ After disconnecting, A $600 \mathrm{pF}$ capacitor added. Then equivalent capacitance $\left(\mathrm{C}^{\prime}\right)$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $=\frac{1}{600}+\frac{1}{600}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{300}$ $\mathrm{C}^{\prime} =300 \mathrm{pF}$ Then, new electrostatic energy, $\mathrm{E}^{\prime} =\frac{1}{2} \mathrm{C}^{\prime} \times \mathrm{V}^{2}$ $=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2}$ $=0.6 \times 10^{-5} \mathrm{~J}$ Loss in electrostatic energy $\left(\mathrm{E}-\mathrm{E}^{\prime}\right)$ $=1.2 \times 10^{-5}-0.6 \times 10^{5}$ $=0.6 \times 10^{-5}$ $=6 \times 10^{-6} \mathrm{~J}$
165700
Two capacitors, one $4 \mathrm{pF}$ and the other $6 \mathrm{pF}$, connected in parallel, are charged by a $100 \mathrm{~V}$ battery. The energy stored in the capacitors is
1 $12 \times 10^{-8} \mathrm{~J}$
2 $2.4 \times 10^{-8} \mathrm{~J}$
3 $5.0 \times 10^{-8} \mathrm{~J}$
4 $1.2 \times 10^{-6} \mathrm{~J}$
Explanation:
: Given, $\mathrm{C}_{1}=4 \mathrm{pF}$ and $\mathrm{C}_{2}=6 \mathrm{pF}$ $\text { Voltage }=100 \mathrm{~V}$ Both capacitors connected in parallel. Hence, equivalent capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=4 \mathrm{pF}+6 \mathrm{pF}=10 \mathrm{pF}$ $\mathrm{C}=10 \times 10^{-12} \mathrm{~F}$ Energy stored in the capacitor $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times(100)^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times 10^{4}$ $\mathrm{E} =5.0 \times 10^{-8} \mathrm{~J}$
[Manipal UGET-2017]
Capacitance
165701
Two conductors of the same material have their diameters in the ratio $1: 2$ and their lengths in the ratio $2: 1$. If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be
165702
As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$, then energy given by the battery will be
1 $\frac{1}{2} \mathrm{CV}^{2}$
2 more than $\frac{1}{2} \mathrm{CV}^{2}$
3 less than $\frac{1}{2} \mathrm{CV}^{2}$
4 zero
Explanation:
: We know, Energy stored in a capacitor $\mathrm{E}=\frac{1}{2} \mathrm{CV}_{\mathrm{C}}^{2}$ Where, $\mathrm{V}_{\mathrm{C}}=$ Voltage drop across capacitor. Apply KVL in Circuit $\mathrm{V}=\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{C}}$ We know, in steady state current through resistance $=0$ $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}=0$ Hence, $\mathrm{V}=\mathrm{V}_{\mathrm{C}}$ From equation (i) $\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}$
[CG PET -2018]
Capacitance
165703
The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by
: We know, Total energy Stored in a capacitor, $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $\text { or } \quad \mathrm{E} =\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}^{2} \quad\left(\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right)$ Volume between the plate $=\mathrm{A} \times \mathrm{d}$ Hence, Energy per unit Volume, $\frac{E}{A d}=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right) V^{2}}{A d}=\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}$
[CG PET- 2017
Capacitance
165704
A $600 \mathrm{pF}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is
1 $6 \times 10^{-6} \mathrm{~J}$
2 $3 \times 10^{-6} \mathrm{~J}$
3 $6 \times 10^{-9} \mathrm{~J}$
4 $3 \times 10^{-9} \mathrm{~J}$
Explanation:
: Given that, Capacitance of Capacitor $(\mathrm{C})=600 \mathrm{pF}=600 \times 10^{-12} \mathrm{~F}$ and potential difference $(\mathrm{V})=200 \mathrm{~V}$ Electrostatic Energy stored $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=\frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^{4}$ $=1.2 \times 10^{-5} \mathrm{~J}$ After disconnecting, A $600 \mathrm{pF}$ capacitor added. Then equivalent capacitance $\left(\mathrm{C}^{\prime}\right)$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $=\frac{1}{600}+\frac{1}{600}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{300}$ $\mathrm{C}^{\prime} =300 \mathrm{pF}$ Then, new electrostatic energy, $\mathrm{E}^{\prime} =\frac{1}{2} \mathrm{C}^{\prime} \times \mathrm{V}^{2}$ $=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2}$ $=0.6 \times 10^{-5} \mathrm{~J}$ Loss in electrostatic energy $\left(\mathrm{E}-\mathrm{E}^{\prime}\right)$ $=1.2 \times 10^{-5}-0.6 \times 10^{5}$ $=0.6 \times 10^{-5}$ $=6 \times 10^{-6} \mathrm{~J}$
165700
Two capacitors, one $4 \mathrm{pF}$ and the other $6 \mathrm{pF}$, connected in parallel, are charged by a $100 \mathrm{~V}$ battery. The energy stored in the capacitors is
1 $12 \times 10^{-8} \mathrm{~J}$
2 $2.4 \times 10^{-8} \mathrm{~J}$
3 $5.0 \times 10^{-8} \mathrm{~J}$
4 $1.2 \times 10^{-6} \mathrm{~J}$
Explanation:
: Given, $\mathrm{C}_{1}=4 \mathrm{pF}$ and $\mathrm{C}_{2}=6 \mathrm{pF}$ $\text { Voltage }=100 \mathrm{~V}$ Both capacitors connected in parallel. Hence, equivalent capacitance, $\mathrm{C}=\mathrm{C}_{1}+\mathrm{C}_{2}=4 \mathrm{pF}+6 \mathrm{pF}=10 \mathrm{pF}$ $\mathrm{C}=10 \times 10^{-12} \mathrm{~F}$ Energy stored in the capacitor $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times(100)^{2}$ $=\frac{1}{2} \times 10 \times 10^{-12} \times 10^{4}$ $\mathrm{E} =5.0 \times 10^{-8} \mathrm{~J}$
[Manipal UGET-2017]
Capacitance
165701
Two conductors of the same material have their diameters in the ratio $1: 2$ and their lengths in the ratio $2: 1$. If the temperature difference between their ends is the same then the ratio of amounts of heat conducted per second through them will be
165702
As shown in the figure below, if a capacitor $C$ is charged by connecting it resistance $R$, then energy given by the battery will be
1 $\frac{1}{2} \mathrm{CV}^{2}$
2 more than $\frac{1}{2} \mathrm{CV}^{2}$
3 less than $\frac{1}{2} \mathrm{CV}^{2}$
4 zero
Explanation:
: We know, Energy stored in a capacitor $\mathrm{E}=\frac{1}{2} \mathrm{CV}_{\mathrm{C}}^{2}$ Where, $\mathrm{V}_{\mathrm{C}}=$ Voltage drop across capacitor. Apply KVL in Circuit $\mathrm{V}=\mathrm{V}_{\mathrm{R}}+\mathrm{V}_{\mathrm{C}}$ We know, in steady state current through resistance $=0$ $\mathrm{V}_{\mathrm{R}}=\mathrm{IR}=0$ Hence, $\mathrm{V}=\mathrm{V}_{\mathrm{C}}$ From equation (i) $\mathrm{E}=\frac{1}{2} \mathrm{CV}^{2}$
[CG PET -2018]
Capacitance
165703
The energy per unit volume for a capacitor having area $A$ and separation $d$ kept at potential different $V$ is given by
: We know, Total energy Stored in a capacitor, $\mathrm{E} =\frac{1}{2} \mathrm{CV}^{2}$ $\text { or } \quad \mathrm{E} =\frac{1}{2}\left(\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right) \mathrm{V}^{2} \quad\left(\because \mathrm{C}=\frac{\varepsilon_{0} \mathrm{~A}}{\mathrm{~d}}\right)$ Volume between the plate $=\mathrm{A} \times \mathrm{d}$ Hence, Energy per unit Volume, $\frac{E}{A d}=\frac{\frac{1}{2}\left(\frac{\varepsilon_{0} A}{d}\right) V^{2}}{A d}=\frac{1}{2} \frac{\varepsilon_{0} V^{2}}{d^{2}}$
[CG PET- 2017
Capacitance
165704
A $600 \mathrm{pF}$ capacitor is charged by a $200 \mathrm{~V}$ supply. It is then disconnected from the supply and is connected to another uncharged $600 \mathrm{pF}$ capacitor. Electrostatic energy lost in the process is
1 $6 \times 10^{-6} \mathrm{~J}$
2 $3 \times 10^{-6} \mathrm{~J}$
3 $6 \times 10^{-9} \mathrm{~J}$
4 $3 \times 10^{-9} \mathrm{~J}$
Explanation:
: Given that, Capacitance of Capacitor $(\mathrm{C})=600 \mathrm{pF}=600 \times 10^{-12} \mathrm{~F}$ and potential difference $(\mathrm{V})=200 \mathrm{~V}$ Electrostatic Energy stored $(\mathrm{E})=\frac{1}{2} \mathrm{CV}^{2}$ $=\frac{1}{2} \times\left(600 \times 10^{-12}\right) \times(200)^{2}$ $=\frac{1}{2} \times 600 \times 10^{-12} \times 4 \times 10^{4}$ $=1.2 \times 10^{-5} \mathrm{~J}$ After disconnecting, A $600 \mathrm{pF}$ capacitor added. Then equivalent capacitance $\left(\mathrm{C}^{\prime}\right)$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{\mathrm{C}}+\frac{1}{\mathrm{C}}$ $=\frac{1}{600}+\frac{1}{600}$ $\frac{1}{\mathrm{C}^{\prime}} =\frac{1}{300}$ $\mathrm{C}^{\prime} =300 \mathrm{pF}$ Then, new electrostatic energy, $\mathrm{E}^{\prime} =\frac{1}{2} \mathrm{C}^{\prime} \times \mathrm{V}^{2}$ $=\frac{1}{2} \times 300 \times 10^{-12} \times(200)^{2}$ $=0.6 \times 10^{-5} \mathrm{~J}$ Loss in electrostatic energy $\left(\mathrm{E}-\mathrm{E}^{\prime}\right)$ $=1.2 \times 10^{-5}-0.6 \times 10^{5}$ $=0.6 \times 10^{-5}$ $=6 \times 10^{-6} \mathrm{~J}$
